Question

Solve the following equation by reduction method:
$x+3y+3z=16$
$x+4y+4z=21$
$x+3y+4z=19$

Answer 1

$\left|\begin{array}{cccc}1& 3& 3& -16\\ 1& 4& 4& -21\\ 1& 3& 4& -19\end{array}\right|$

$R_2\to R_2-R_1$

$=\left|\begin{array}{cccc}1& 3& 3& -16\\ 0& 1& 0& -5\\ 1& 3& 4& -19\end{array}\right|$

$R_3\to R_3-R_1$

$=\left|\begin{array}{cccc}1& 3& 3& -16\\ 0& 1& 0& -5\\ 0& 0& 1& -3\end{array}\right|$

$\Rightarrow z=-3,y=-5,x+3y+3z=-16$

$\Rightarrow x+3\times -5+3\times -3=-16$

$\Rightarrow x-15-9=-16$

$\Rightarrow x=-16+24=8$

$\therefore x=8,y=-5,z=-3$